Linear Combinations and Span

Let v1, v2,…, vrbe vectors in Rn. A linear combination of these vectors is any expression of the form

where the coefficients k1, k2,…, k rare scalars.

Example 1: The vector v = (−7, −6) is a linear combination of the vectors v1 = (−2, 3) and v2 = (1, 4), since v = 2 v1 − 3 v2. The zero vector is also a linear combination of v1 and v2, since 0 = 0 v1 + 0 v2. In fact, it is easy to see that the zero vector in Rn is always a linear combination of any collection of vectors v1, v2,…, vrfrom Rn.

The set of all linear combinations of a collection of vectors v1, v2,…, vrfrom Rn is called the span of { v1, v2,…, vr}. This set, denoted span { v1, v2,…, vr}, is always a subspace of Rn, since it is clearly closed under addition and scalar multiplication (because it contains all linear combinations of v1, v2,…, vr). If V = span { v1, v2,…, vr}, then V is said to be spanned by v1, v2,…, vr.

Example 2: The span of the set {(2, 5, 3), (1, 1, 1)} is the subspace of R3 consisting of all linear combinations of the vectors v1 = (2, 5, 3) and v2 = (1, 1, 1). This defines a plane in R3. Since a normal vector to this plane in n = v1 x v2 = (2, 1, −3), the equation of this plane has the form 2 x + y − 3 z = d for some constant d. Since the plane must contain the origin—it's a subspace— d must be 0. This is the plane in Example 7.

Example 3: The subspace of R2 spanned by the vectors i = (1, 0) and j = (0, 1) is all of R2, because every vector in R2 can be written as a linear combination of i and j:

That is, if any one of the vectors in a given collection is a linear combination of the others, then it can be discarded without affecting the span. Therefore, to arrive at the most “efficient” spanning set, seek out and eliminate any vectors that depend on (that is, can be written as a linear combination of) the others.

That is, because v3 is a linear combination of v1 and v2, it can be eliminated from the collection without affecting the span. Geometrically, the vector (3, 15, 7) lies in the plane spanned by v1 and v2 (see Example 7 above), so adding multiples of v3 to linear combinations of v1 and v2 would yield no vectors off this plane. Note that v1 is a linear combination of v2 and v3 (since v1 = 5/4 v2 + 1/4 v3), and v2 is a linear combination of v1 and v3 (since v2 = 4/5 v1 − 1/5 v3). Therefore, any one of these vectors can be discarded without affecting the span:

Example 5: Let v1 = (2, 5, 3), v2 = (1, 1, 1), and v3 = (4, −2, 0). Because there exist no constants k1 and k2 such that v3 = k1v1 + k2v2, v3 is not a linear combination of v1 and v2. Therefore, v3 does not lie in the plane spanned by v1 and v2, as shown in Figure :

Figure 1

Consequently, the span of v1, v2, and v3 contains vectors not in the span of v1 and v2 alone. In fact,